Grothendieck with an exact outer functor

Applying an exact $G$ to any augmented injective resolution preserves its positive exactness, so $R^{p}G(V)=0$ for $p>0$ and every $V$ is $G$-acyclic. The Grothendieck injective-image condition is therefore automatic. The page has only column zero; for every $r\ge 2$ outgoing differentials land in a zero positive column and incoming ones start in a negative column. Consequently $E_2=E_{\mathrm{\infty }}$, and $F^1{H}^n=0,F^0{H}^n=H^n$ reconstruct the target through the upper edge in F1.

For the displayed specialization resolve $\mathbb{Z}\mathrm{/}2$ by $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$. The free rank-one terms are projective by lifting the image of $1$. Applying $\mathrm{Hom}(-,\mathbb{Z})$ gives $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$, with kernel zero and cokernel $\mathbb{Z}\mathrm{/}2$. F2 thus gives $R^{0}F(\mathbb{Z})=0$, $R^{1}F(\mathbb{Z})=\mathbb{Z}\mathrm{/}2$ and all higher terms zero. The target is zero outside degree one and is $\mathbb{Z}\mathrm{/}2$ in degree one; its upper edge is the identity under these common Hom-cohomology identifications. The lower edge is zero in positive degrees because its source is a positive derived identity functor. No extension or additional splitting choice remains.